Let the radius of the sphere be \(r_1\) cm when its curved surface area is \(16\pi\) square cm, and \(r_2\) cm when the curved surface area is \(4\pi\) square cm. \(\therefore 4\pi {r_1}^2 = 16\pi\) i.e., \({r_1}^2 = 4\) i.e., \(r_1 = 2\) cm And \(4\pi {r_2}^2 = 4\pi\) i.e., \({r_2}^2 = 1\) i.e., \(r_2 = 1\) cm \(\therefore\) Percentage decrease in volume of the sphere \(= \frac{\frac{4}{3}\pi 2^3 - \frac{4}{3}\pi 1^3}{\frac{4}{3}\pi 2^3} \times 100\%\) \(= \frac{8 - 1}{8} \times 100\%\) \(= \frac{175}{2}\% = 87\frac{1}{2}\%\)